I have to solve a system of non-linear equations of the following function:
function eq = ts_7(A,B,C,D,E)
syms x
% dbstop if error
g = D * sin(C * atan( B*x - E * (B*x - atan(B*x)))) + A; % magic formula
eq5 = taylor(g, x, 'Order',1, 'ExpansionPoint',7) + 4296; % x^0
eq1 = taylor(g, x, 'Order',2, 'ExpansionPoint',7) - eq5 + 296.3; % x^1
eq2 = taylor(g, x, 'Order',3, 'ExpansionPoint',7) - eq1 - 79.77; % x^2
eq3 = taylor(g, x, 'Order',4, 'ExpansionPoint',7) - eq2 - 4.541; % x^3
eq4 = taylor(g, x, 'Order',5, 'ExpansionPoint',7) - eq3 - 0.03358; % x^4
eq{1} = matlabFunction(eq1) % syms to numeric function
eq{2} = matlabFunction(eq2)
eq{3} = matlabFunction(eq3)
eq{4} = matlabFunction(eq4)
eq{5} = matlabFunction(eq5)
end
and my main is the following
opts = optimoptions('fsolve','InitDamping',0.005,'Algorithm','levenberg-marquardt');
init = [-1.3, 1.4, 4000, 0.12, 9]; % starting points
tic
coeff = fsolve(#(x)ts_7(x(1), x(2), x(3), x(4), x(5)), init,opts);
toc
and the error I keep getting is
Undefined function or variable "fuser".
Error in fsolve (line 257)
if ~isempty( isoptimargdbl('FSOLVE', {'F','J'}, fuser, JAC) )
Error in script_7 (line 11)
coeff = fsolve(#(x)ts_7(x(1), x(2), x(3), x(4), x(5)), init,opts);
Error in run (line 96)
evalin('caller', [script ';']);
I don't know how to fix it. I also tried solving the same problem with vpasolve and solve. For solve, it takes too long and for vpasolve, I get the error that symbolic parameters are not allowed in a non-poly equation.
Is there a way to transform string into a function (not a handle function what matlabFunction is doing in this code)?
I am getting desperate because I don't want to rewrite everything by hand.
UPDATE: I tried solving 2-D system, example from http://nl.mathworks.com/help/optim/ug/fsolve.html. I get the same error, here is the screenshot.
Your function ts_7 needs to return a floating point vector, but eq is a cell array of function handles. You also don't need to call matlabFunction five time. Instead:
eq = matlabFunction([eq1;eq2;eq3;eq4;eq5]);
Now eq is a function handle that returns a 5-by-1 column vector of doubles. This needs to be evaluated numerically at the expansion point x so that the output of ts_7 is a floating point vector, for example:
eq = eq(1);
It's not clear to me if you're also optimizing x or if you know this.
There was something wrong with my MATLAB. I solved the problem by using another version of MATLAB.
Related
I have written the following Matlab function:
function EulerMethod(t_min,t_max,h,f,Y,yzero)
tlist = t_min:h:t_max;
N = (t_max - t_min)/h;
ylist = transpose(zeros(N+1,1));
ylist(1) = yzero;
for i=1:N
term = f(tlist(i),ylist(i))*h;
ylist(i+1) = ylist(i) + term;
end
yrange = Y(tlist);
% modified to generate a new figure window each time.
figure;
plot(tlist,yrange,'red','LineWidth', 2);
hold;
plot(tlist,ylist,'blue','LineWidth', 2);
plot(tlist, abs(yrange - ylist),'magenta','LineWidth', 2)
% modified to wrap the title.
title({'Graphs of the True Solution, Euler Solution,', 'and the Absolute Value of the Global Error (GE)'})
xlabel('t')
ylabel('Y(t)')
legend({'True Solution','Euler Solution', 'Absolute Value of the GE'},'Location','southwest')
end
I now try and call this function in another script. The variables f and Y in the function are symbolic functions; so, in the other file, I first declare these functions before calling this function. Here's the code:
clc
syms f(t,y)
syms Y(t)
f(t,y) = -y + 2.0*cos(t); %This the derivative of the function whose solution we're trying to computed
Y(t) = sin(t) + cos(t); %This is the true solution;
% Calling function
EulerMethod(0.0, 6.0, 0.2, f, Y, 1.0);
I, however, get errors when I run the second script. Can anyone help me figure out what's going wrong? I suspect this may be because of the way I have both input and used the symbolic functions f and Y but I am not sure.
Consider the following MWE in MATLAB:
f = #(t) integral(#(x) x.^2,0,t);
integral(f,0,1);
This yields the error
Error using integral (line 85) A and B must be floating-point scalars.
(and a bit more). How do I fix this? Is this even possible? I think the problem is the variable upper bound.
If you want to use integral then set 'ArrayValued' to true otherwise t would be an invalid end point in integral(#(x) x.^2,0,t). So it would be:
f = #(t) integral(#(x) x.^2,0,t);
integral(f,0,1,'ArrayValued',true)
% ans =
% 0.0833
Alternately, since you're doing double integration, so use the function dedicated for this purpose i.e. integral2. For your example, it would be:
f = #(t,x) x.^2 ;
integral2(f,0,1,0, #(t) t)
% ans =
% 0.0833
If you have Symbolic Math Toolbox, you can also use int as int(expr,var,a,b) but it would be slower. For your case, it would be:
syms x t;
f = x.^2;
req = int(int(f,x,0,t),t,0,1); % It gives 1/12
req = double(req); % Convert to double if required
I am trying to implement the Taylor method for ODEs in MatLab:
My code (so far) looks like this...
function [x,y] = TaylorEDO(f, a, b, n, y0)
% syms t
% x = sym('x(t)'); % x(t)
% f = (t^2)*x+x*(1-x);
h = (b - a)/n;
fprime = diff(f);
f2prime = diff(fprime);
y(0) = y0,
for i=1:n
T((i-1)*h, y(i-1), n) = double(f((i-1)*h, y(i-1)))+(h/2)*fprime((i-1)*h, y(i-1))
y(i+1) = w(i) + h*T(t(i), y(i), n);
I was trying to use symbolic variables, but I don´t know if/when I have to use double.
I also tried this other code, which is from a Matlab function, but I do not understand how f should enter the code and how this df is calculated.
http://www.mathworks.com/matlabcentral/fileexchange/2181-numerical-methods-using-matlab-2e/content/edition2/matlab/chap_9/taylor.m
As error using the function from this link, I got:
>> taylor('f',0,2,0,20)
Error using feval
Undefined function 'df' for input arguments of type 'double'.
Error in TaylorEDO (line 28)
D = feval('df',tj,yj)
The f I used here was
syms t
x = sym('x(t)'); % x(t)
f = (t^2)*x+x*(1-x);
This is a numerical method, so it needs numerical functions. However, some of them are computed from the derivatives of the function f. For that, you need symbolic differentiation.
Relevant Matlab commands are symfun (create a symbolic function) and matlabFunction (convert a symbolic function to numerical).
The code you have so far doesn't seem salvageable. You need to start somewhere closer to basics, e.g., "Matlab indices begin at 1". So I'll fill the gap (computation of df) in the code you linked to. The comments should explain what is going on.
function [T,Y] = taylor(f,a,b,ya,m)
syms t y
dfs(1) = symfun(f, [t y]); % make sure that the function has 2 arguments, even if the user passes in something like 2*y
for k=1:3
dfs(k+1) = diff(dfs(k),t)+f*diff(dfs(k),y); % the idea of Taylor method: calculate the higher derivatives of solution from the ODE
end
df = matlabFunction(symfun(dfs,[t y])); % convert to numerical function; again, make sure it has two variables
h = (b - a)/m; % the rest is unchanged except one line
T = zeros(1,m+1);
Y = zeros(1,m+1);
T(1) = a;
Y(1) = ya;
for j=1:m
tj = T(j);
yj = Y(j);
D = df(tj,yj); % syntax change here; feval is unnecessary with the above approach to df
Y(j+1) = yj + h*(D(1)+h*(D(2)/2+h*(D(3)/6+h*D(4)/24)));
T(j+1) = a + h*j;
end
end
Example of usage:
syms t y
[T, Y] = taylor(t*y, 0, 1, 2, 100);
plot(T,Y)
How can I use matlab to solve the following Ordinary Differential Equations?
x''/y = y''/x = -( x''y + 2x'y' + xy'')
with two known points, such as t=0: x(0)= x0, y(0) = y0; t=1: x(1) = x1, y(1) = y1 ?
It doesn't need to be a complete formula if it is difficult. A numerical solution is ok, which means, given a specific t, I can get the value of x(t) and y(t).
If matlab is hard to do this, mathematica is also OK. But as I am not familiar with mathematica, so I would prefer matlab if possible.
Looking forward to help, thanks!
I asked the same question on stackexchange, but haven't get good answer yet.
https://math.stackexchange.com/questions/812985/matlab-or-mathematica-solve-ordinary-differential-equations
Hope I can get problem solved here!
What I have tried is:
---------MATLAB
syms t
>> [x, y] = dsolve('(D2x)/y = -(y*D2x + 2Dx*Dy + x*D2y)', '(D2y)/x = -(y*D2x + 2Dx*Dy + x*D2y)','t')
Error using sym>convertExpression (line 2246)
Conversion to 'sym' returned the MuPAD error: Error: Unexpected 'identifier'.
[line 1, col 31]
Error in sym>convertChar (line 2157)
s = convertExpression(x);
Error in sym>convertCharWithOption (line 2140)
s = convertChar(x);
Error in sym>tomupad (line 1871)
S = convertCharWithOption(x,a);
Error in sym (line 104)
S.s = tomupad(x,'');
Error in dsolve>mupadDsolve (line 324)
sys = [sys_sym sym(sys_str)];
Error in dsolve (line 186)
sol = mupadDsolve(args, options);
--------MATLAB
Also, I tried to add conditions, such as x(0) = 2, y(0)=8, x(1) = 7, y(1) = 18, and the errors are still similar. So what I think is that this cannot be solve by dsolve function.
So, again, the key problem is, given two known points, such as when t=0: x(0)= x0, y(0) = y0; t=1: x(1) = x1, y(1) = y1 , how I get the value of x(t) and y(t)?
Update:
I tried ode45 functions. First, in order to turn the 2-order equations into 1-order, I set x1 = x, x2=y, x3=x', x4=y'. After some calculation, the equation becomes:
x(1)' = x(3) (1)
x(2)' = x(4) (2)
x(3)' = x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)) (3)
x(4)' = -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2) (4)
So the matlab code I wrote is:
myOdes.m
function xdot = myOdes(t,x)
xdot = [x(3); x(4); x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)); -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)]
end
main.m
t0 = 0;
tf = 1;
x0 = [2 3 5 7]';
[t,x] = ode45('myOdes',[t0,tf],x0);
plot(t,x)
It can work. However, actually this is not right. Because, what I know is that when t=0, the value of x and y, which is x(1) and x(2); and when t=1, the value of x and y. But the ode functions need the initial value: x0, I just wrote the condition x0 = [2 3 5 7]' randomly to help this code work. So how to solve this problem?
UPDATE:
I tried to use the function bvp4c after I realized that it is a boundary value problem and the following is my code (Suppose the two boundry value conditions are: when t=0: x=1, y=3; when t=1, x=6, y=9. x is x(1), y is x(2) ):
1. bc.m
function res = bc(ya,yb)
res = [ ya(1)-1; ya(2)-3; yb(1) - 6; yb(2)-9];
end
2. ode.m
function dydx = ode(t,x)
dydx = [x(3); x(4); x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)); -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)];
end
3. mainBVP.m
solinit = bvpinit(linspace(0,6,10),[1 0 -1 0]);
sol = bvp4c(#ode,#bc,solinit);
t = linspace(0,6);
x = deval(sol,t);
plot(t,x(1,:));
hold on
plot(t,x(2,:));
plot(t,x(3,:));
plot(t,x(4,:));
x(1,:)
x(2,:)
It can work, but I don't know whether it is right. I will check it again to make sure it is the right code.
As mentioned, this isn't a math site, so try to give code or something showing some effort.
However, the first step you need to do is turn the DE into normal form (i.e., no 2nd derivatives). You do this by making a separate variable equal to the derivative. Then, you use
syms x y % or any variable instead of x or y
to define variables as symbolic. Use matlabfunction to create a symbolic function based on these variables. Finally, you can use the ode45 function to solve the symbolic function while passing variable values. I recommend you look up the full documentation in matlab in order to understand it better, but here is a very basic syntax:
MyFun= matlabFunction(eq,'vars',{x,y});
[xout,yout]=ode45(#(x,Y) MyFun(variables),[variable values],Options);
Hopefully this puts you in the right direction, so try messing around with it and provide code if you need more help.
EDIT:
This is how I would solve the problem. Note: I don't really like the matlabFunction creator but this is simply a personal preference for various reasons I won't go into.
% Seperate function of the first order state equations
function dz = firstOrderEqns(t,z)
dz(4,1) = 0;
dz(1) = -2.*z(3).*z(1).*z(4)./(1 + z(4).^2 + z(2).^2);
dz(2) = z(1);
dz(3) = -2.*z(2).*z(3).*z(1)./(1 + z(4).^2 + z(2).^2);
dz(4) = z(3);
end
% runfirstOrderEqns
%% Initial conditions i.e. # t=0
z1 = 5; % dy/dt = 5 (you didn't specify these initial conditions,
% these will depend on the system which you didn't really specify
z2 = 0; % y = 0
z3 = 5; % dx/dt = 5 (The same as for z1)
z4 = 0; % x = 0
IC = [z1, z2, z3, z4];
%% Run simulation
% Time vector: i.e closed interval [0,20]
t = [0,20]; % This is where you have to know about your system
% i.e what is it's time domain.
% Note: when a system has unstable poles at
% certain places the solver can crash you need
% to understand these.
% using default settings (See documentation ode45 for 'options')
[T,Y] = ode45(#firstOrderEqns,t,IC);
%% Plot function
plot(T,Y(:,1),'-',T,Y(:,2),'-.',T,Y(:,3),':',T,Y(:,4),'.');
legend('dy/dt','y','dx/dt','x')
As in my comments I have made a lot of assumtions that you need to fix for example, you didn't specify what the initial conditions for the first derivatives of the states are i.e. (z1, z3) which is important for the response of the system. Also you didn't specify the time interval your interested for the simulation etc.
Note: The second m file can be used with any state function in the correct format
The following is the answer we finally get #Chriso: use matlab bvp4c function to solve this boundary value problem (Suppose the two boundry value conditions are: when t=0: x=1, y=3; when t=1, x=6, y=9. x is x(1), y is x(2) ):
1. bc.m
function res = bc(ya,yb)
res = [ ya(1)-1; ya(2)-3; yb(1) - 6; yb(2)-9];
end
2. ode.m
function dydx = ode(t,x)
dydx = [x(3); x(4); x(2)/x(1)*(-2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)); -2*x(1)*x(3)*x(4)/(1+x(1)^2+x(2)^2)];
end
3. mainBVP.m
solinit = bvpinit(linspace(0,6,10),[1 0 -1 0]);
sol = bvp4c(#ode,#bc,solinit);
t = linspace(0,6);
x = deval(sol,t);
plot(t,x(1,:));
hold on
plot(t,x(2,:));
plot(t,x(3,:));
plot(t,x(4,:));
x(1,:)
x(2,:)
So I need to solve x''(t) = -x(t)^p with initial conditions x(0)= 0 and v(0) = x'(0) = v_o = 1.
The value of the parameter p is 1.
This is what I have:
function [t, velocity, x] = ode_oscilation(p)
y=[0;0;0];
% transform system to the canonical form
function y = oscilation_equation(x,p)
y=zeros(2,1);
y(1)=y(2);
y(2)=-(x)^p;
% to make matlab happy we need to return a column vector
% so we transpose (note the dot in .')
y=y.';
end
tspan=[0, 30]; % time interval of interest
[t,velocity,x] = ode45(#oscilation_equation, tspan, 1);
t = y(:,1);
xposition=y(:,3);
velocity=y(:,2);
end
and this is the error message I receive:
ode_oscillation(1)
Error using odearguments (line 91)
ODE_OSCILLATION/OSCILATION_EQUATION must return a
column vector.
Error in ode45 (line 114)
[neq, tspan, ntspan, next, t0, tfinal, tdir, y0, f0,
odeArgs, odeFcn, ...
Error in ode_oscillation (line 17)
[t,velocity,x] = ode45(#oscilation_equation, tspan,1);
There's a few things going wrong here. First, from help ode45:
ode45 Solve non-stiff differential equations, medium order method.
[TOUT,YOUT] = ode45(ODEFUN,TSPAN,Y0) with TSPAN = [T0 TFINAL] integrates
the system of differential equations y' = f(t,y) from time T0 to TFINAL
with initial conditions Y0.
Note that ode45 expects a function f(t,y), where size(t) == [1 1] for time and size(y) == [1 N] or [N 1] for solution values. Your oscilation_equation has the order of input arguments inverted, and you input a constant parameter p instead of time t.
Also, the initial conditions Y0 should have the same size as y; so size(y0) == [N 1] or [1 N]. You just have 1, which is clearly causing errors.
Also, your output arguments t, xposition and velocity will be completely ignored and erroneous, since y is not set as output argument from ode45, and most of all, their names do not correspond to ode_oscilation's output arguments. Also, their order of extracting from columns of y is incorrect.
So, in summary, change everything to this:
function [t, v, x] = ode_oscilation(p)
% initial values
y0 = [0 1];
% time interval of interest
tspan =[0 30];
% solve system
[t,y] = ode45(#(t,y) [y(2); -y(1)^p], tspan, y0);
% and return values of interest
x = y(:,1);
v = y(:,2);
end